Skip to content

3561. Resulting String After Adjacent Removals 👍

  • Time: $O(n)$
  • Space: $O(1)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
class Solution {
 public:
  string resultingString(string s) {
    string ans;

    for (const char c : s)
      if (!ans.empty() &&
          (abs(ans.back() - c) == 1 || abs(ans.back() - c) == 25))
        ans.pop_back();
      else
        ans += c;

    return ans;
  }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
class Solution {
  public String resultingString(String s) {
    StringBuilder sb = new StringBuilder();

    for (final char c : s.toCharArray())
      if (sb.length() > 0 && (Math.abs(sb.charAt(sb.length() - 1) - c) == 1 ||
                              Math.abs(sb.charAt(sb.length() - 1) - c) == 25))
        sb.deleteCharAt(sb.length() - 1);
      else
        sb.append(c);

    return sb.toString();
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
class Solution:
  def resultingString(self, s: str) -> str:
    stack = []

    for c in s:
      if stack and abs(ord(stack[-1]) - ord(c)) in (1, 25):
        stack.pop()
      else:
        stack.append(c)

    return ''.join(stack)